last two digits of $9^{1500}$ (Dummit Foote -Abstract Algebra preliminaries $0.3.5$)

You have seen that $9^2 \equiv 9^{32} \equiv 81$ mod $100$. As gcd$(81,100) = 1$, this implies $9^{30} \equiv \frac{9^{32}}{9^2} \equiv \frac{81}{81} \equiv 1$, and thus $9^{1500} = (9^{30})^{50} \equiv 1$.

Your method also works, but it is longer.

$$ \phi(100)=\phi(5^2)\phi(2^2)=40 $$ where $\phi$ is Euler's Totient function. Then $$ 9^{1500}=9^{20}9^{37*40}=9^{20}=3^{40}=1 $$ Using Fermat's little theorem.

A similar question has been asked a while back. The whole idea is to exploit the fact that
$9^n=(10-1)^n$ and/or the fact that $9^{2n}=81^n=(8\cdot10+1)^n$, and plug it into Newton's
Binomial Theorem. It will soon become evident that our number is of the form $\mathcal{M}_{100}+1$.